ThatQuiz Elenco di test Affronta questo test adesso
ALG: Exam review Ch 11 - growth/decay
Con il contributo di: Maynard
(Autore originale: Crouch)
y = 30(0.6x)
y = (⅞)x

Tell whether the following represent exponential

growth or decay.  Type the word growth or decay. 

Spelling counts!

y = 2(5)x

A new car that sells for $31,000 and is losing it's

value at a rate of 23% each year.   Write a function

that models the value of the car.  Simplify the

number in parenthesis. 

Do NOT use any commas, $, or extra zeros in your

answer.

y=
(
)
x

Suppose you invest $2000 at an annual rate of

5% compounded weekly.  How much will you have

in the account after 4 yrs?

  $2431.01

 $2442.57

Formula:  A = P(1+ )tn

                                      n

 $10,125.00

 $400.00

f(x) = (712 + 0.24)x

 f(x) =712(1.24)x

An initial population of 712 quail increases at an

 annual rate of 24%.  Write an exponential function

to model the quail population.

 

 f(x) =  712(0.24)x

 f(x) = 712(0.76)x

y =P(.5)t

The half-life of an element is 6 hours.

If you have 56mg of the element, how much

will you have after 15 hours?

 12mg

 20mg

 5.6mg

 9.9mg
y=8x
y =1.2(3)x

In the following exponential functions,

what are the values of "a" and "b"?

b:

a:
b:
a: 

Mrs. Cooke is buying a painting for $1,500.  (She

found a money tree in her closet!) It appreciates

(grows) in value 22% every year.  Write a function

that models the value of the painting.  Simplify the

number in parenthesis.

Do NOT use any commas, $, or extra zeros in your

answer.

y=
(
)
x
y =P(.5)t

The half-life of a Iodine-32 is 2 days.

If you have 100-mg of Iodine-32, how much

will you have after 12 days?

.5 mg

 .003 mg

 .02 mg

 1.56 mg

A new car that sells for $31,000 and is losing it's

value at a rate of 23% each year.   What is the value

of the car in 5 years?

$8391.03

$87,274.48

$27,595.26

$19.95

In the year 2000, the population in Canada was

about 85 million and growing at a rate of

1.32% each year.  At this growth rate, the

function f(x) = 85(1.0132)x gives the population,

in millions, x years after 2000.  Using this model,

in what year will the population reach 100 million?

2011

2012

2013

2014

 

The equation f(x) = 3,150(.95)x represents the

declining populations of cardinals in one Michigan

county.  How many cardinals will there be in 7 years?

20,948 cardinals

2200 cardinals

2.1 x 1024 cardinals

2395 cardinals

14 rabbits

12 rabbits

16 rabbits

18 rabbits

The rabbit population that is in the back of Mrs.

Maynard's house is growing at a rate of 18%

per year.  If there are 8 rabbits trying to get into

her garden in 2012, how many rabbits will be

trying to get in in 2016?

(round to the nearest whole rabbit)
Studenti hanno provato anche :

Creato con That Quiz — dove la realizzazione e l’esecuzione di test sono resi semplici per la matematica e per altre aree disciplinari.